Question #ab4db

Yes, you are right. The quadratic formula is one way to solve this particular equation. Alternatively, you can also solve by graphing. I won't go into too much detail on the actual process of solving this equation by graphing, but I will just say that you look where the graph intersects the #x#-axis (these would be the solution(s)).

If we use the quadratic formula, we will have to associate the equation given with

#ax^2 + bx + c = 0#.

If we substitute our given equation #x^2-10x+40=0# into the above equation, we get:

#a=1#

#b=-10#

#c=40#

These values give us just enough information to use the quadratic formula, which is:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

So if we substitute our values that we found earlier into this unique formula, we will get #x#.

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

#x = (-(-10) +- sqrt((-10)^2 - 4(1)(40)))/(2(1))#

#x=(10 +- sqrt(100-160))/(2)#

#x=(10+- sqrt(-60))/2#

At this point, you will see a negative square root, which indicates that the solution is no real numbers. But if you are in a higher level of math, and you have learned about imaginary numbers (like #i#), then you are not yet finished with the question. So, let's continue.

From here, we simplify the radical using the imaginary number #i# (#i=sqrt(-1)#). See below.

#x=(10+- sqrt(-60))/2#

#x=(10 +- sqrt(-20 * 2 * 2))/(2)#

#x=(10 +- sqrt(-15 * 2 * 2))/(2)#

#x=(10+-2sqrt(-15))/(2)#

#x=(10+-2isqrt(15))/(2)#

If you want an exact answer (no rounding), then this would be your answer. If you want to put the solutions in #a+bi# form, I leave the task to you, as we are not far off.

I hope that helps!